# Triple Integrals in Spherical Coordinates

The spherical coordinates of a point $$M\left( {x,y,z} \right)$$ are defined to be the three numbers: $$\rho, \varphi, \theta,$$ where

• $$\rho$$ is the length of the radius vector to the point $$M$$;
• $$\varphi$$ is the angle between the projection of the radius vector $$\overrightarrow {OM}$$ on the $$xy$$-plane and the $$x$$-axis;
• $$\theta$$ is the angle of deviation of the radius vector $$\overrightarrow {OM}$$ from the positive direction of the $$z$$-axis (Figure $$1$$).

It's important to take into account that the definition of $$\rho$$ differs in spherical and cylindrical coordinates.

The spherical coordinates of a point are related to its Cartesian coordinates as follows:

$x = \rho \cos \varphi \sin \theta ,\;\;y = \rho \sin \varphi \sin \theta ,\;\; z = \rho \cos \theta ,$
$\text{where}\;\;\rho \ge 0,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .$

The Jacobian of transformation from Cartesian to cylindrical coordinates is written as

$I\left( {\rho ,\varphi ,\theta } \right) = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial \theta }}}\\ {\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial \theta }}}\\ {\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial \theta }}} \end{array}} \right|,$

where the partial derivatives are given by

$\frac{{\partial x}}{{\partial \rho }} = \cos \varphi \sin \theta,\;\; \frac{{\partial x}}{{\partial \varphi }} = - \rho \sin \varphi \sin \theta,\;\; \frac{{\partial x}}{{\partial \theta }} = \rho \cos \varphi \cos \theta,\;\; \frac{{\partial y}}{{\partial \rho }} = \sin \varphi \sin \theta,\;\; \frac{{\partial y}}{{\partial \varphi }} = - \rho \cos \varphi \sin \theta,\;\; \frac{{\partial y}}{{\partial \theta }} = \rho \sin \varphi \cos \theta,\;\; \frac{{\partial z}}{{\partial \rho }} = \cos \theta,\;\; \frac{{\partial z}}{{\partial \varphi }} = 0,\;\; \frac{{\partial z}}{{\partial \theta }} = -\rho \sin \theta .$

By expanding the determinant along the second column, we get

$I\left( {\rho ,\varphi ,\theta } \right) = \rho \sin \varphi \sin \theta \cdot \left| {\begin{array}{*{20}{c}} {\sin \varphi \sin \theta }&{\rho \sin \varphi \cos \theta }\\ {\cos \theta }&{ - \rho \sin \theta } \end{array}} \right| + \rho \cos \varphi \sin \theta \cdot \left| {\begin{array}{*{20}{c}} {\cos \varphi \sin \theta }&{\rho \cos \varphi \cos \theta }\\ {\cos \theta }&{ - \rho \sin \theta } \end{array}} \right| = \rho \sin \varphi \sin \theta \left( { - \rho \sin \varphi \,{{\sin }^2}\theta - \rho \sin \varphi \,{{\cos }^2}\theta } \right) + \rho \cos \varphi \sin \theta \left( { - \rho \cos \varphi \,{{\sin }^2}\theta - \rho \cos \varphi \,{{\cos }^2}\theta } \right) = \rho \sin \varphi \sin \theta \cdot \left( { - \rho \sin \varphi } \right) \cdot 1 + \rho \cos \varphi \sin \theta \cdot \left( { - \rho \cos \varphi } \right) \cdot 1 = - {\rho ^2}\sin \theta \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right) = - {\rho ^2}\sin \theta .$

Accordingly, the absolute value of the Jacobian is

$\left| {I\left( {\rho ,\varphi ,\theta } \right)} \right| = {\rho ^2}\sin \theta .$

Hence, the formula of change of variables for this transformation is

$\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} = \iiint\limits_{U'} {f\left( {{\rho \cos \varphi \sin \theta ,\,\rho \sin \varphi \sin \theta,\, \rho \cos \theta }} \right) {\rho ^2}\sin \theta d\rho d\varphi d\theta } .$

It is easier to calculate triple integrals in spherical coordinates when the region of integration $$U$$ is a ball (or some portion of it) and/or when the integrand is a kind of $$f\left( {{x^2} + {y^2} + {z^2}} \right).$$

It is sometimes more convenient to use so-called generalized spherical coordinates, related to the Cartesian coordinates by the formulas

$x = a\rho \cos \varphi \sin \theta ,\;\; y = b\rho \sin \varphi \sin \theta ,\;\; z = c\rho \cos \theta .$

In this case the Jacobian is

$I\left( {\rho ,\varphi ,\theta } \right) = - abc{\rho ^2}\sin \theta .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Evaluate the integral $\iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz},$ where the region of integration $$U$$ is the ball given by the equation ${x^2} + {y^2} + {z^2} = 25.$

### Example 2

Calculate the integral $\iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} ,$ where the region $$U$$ is the unit ball ${x^2} + {y^2} + {z^2} \le 1.$

### Example 1.

Evaluate the integral $\iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz},$ where the region of integration $$U$$ is the ball given by the equation ${x^2} + {y^2} + {z^2} = 25.$

Solution.

As the region $$U$$ is a ball and the integrand is expressed by a function depending on $$f\left( {{x^2} + {y^2} + {z^2}} \right),$$ we can convert the triple integral to spherical coordinates. Make the substitution:

$x = \rho \cos \varphi \sin \theta ,\;\;y = \rho \sin \varphi \sin \theta ,\;\; z = \rho \cos \theta ,$

The new variables range within the limits:

$0 \le \rho \le 5,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .$

Taking into account the Jacobian $${\rho ^2}\sin \theta,$$ we can write the integral as follows:

$I = \iiint\limits_U {\sqrt {{x^2} + {y^2} + {z^2}} dxdydz} = \iiint\limits_{U'} {\rho \cdot {\rho ^2}\sin \theta d\rho d\varphi d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \int\limits_0^\pi {\sin \theta d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } \left( { - \cos \pi + \cos 0} \right) = 2\int\limits_0^{2\pi } {d\varphi } \int\limits_0^5 {{\rho ^3}d\rho } = 2\int\limits_0^{2\pi } {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^4}}}{4}} \right)} \right|_0^5} \right] = 2\int\limits_0^{2\pi } {d\varphi } \cdot \frac{{{5^4}}}{4} = \frac{{625}}{2}\int\limits_0^{2\pi } {d\varphi } = \frac{{625}}{2} \cdot 2\pi = 625\pi .$

### Example 2.

Calculate the integral $\iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} ,$ where the region $$U$$ is the unit ball ${x^2} + {y^2} + {z^2} \le 1.$

Solution.

This ball is centered at the origin. Hence, the region of integration $$U$$ in spherical coordinates is described by the inequalities

$0 \le \rho \le 1,\;\; 0 \le \varphi \le 2\pi ,\;\; 0 \le \theta \le \pi .$

Writing the integral in spherical coordinates, we have

$I = \iiint\limits_U {{e^{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}}}}dxdydz} = \iiint\limits_{U'} {{e^{{{\left( {{\rho ^2}} \right)}^{\frac{3}{2}}}}}{\rho ^2}\sin \theta d\rho d\varphi d\theta } = \iiint\limits_{U'} {{e^{{\rho ^3}}}{\rho ^2}\sin \theta d\rho d\varphi d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{e^{{\rho ^3}}}{\rho ^2}d\rho } \int\limits_0^\pi {\sin \theta d\theta } .$

As it can be seen, the triple integral is transformed into the product of three single-valued integrals, each of which can be calculated independently. As a result, we obtain

$I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{e^{{\rho ^3}}}{\rho ^2}d\rho } \int\limits_0^\pi {\sin \theta d\theta } = \left[ {\left. \varphi \right|_0^{2\pi }} \right] \cdot \int\limits_0^1 {\left( {{e^{{\rho ^3}}} \cdot \frac{1}{3}d{\rho ^3}} \right)} \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] = 2\pi \cdot \frac{1}{3}\left[ {\left. {\left( {{e^{{\rho ^3}}}} \right)} \right|_{{\rho ^3} = 0}^{{\rho ^3} = 1}} \right] \cdot \left( { - \cos \pi + \cos 0} \right) = \frac{{2\pi }}{3} \cdot \left( {e - 1} \right) \cdot 2 = \frac{{4\pi }}{3}\left( {e - 1} \right).$

See more problems on Page 2.